1. Instrumental Analysis Polarography and Voltammetry
Tutorial 9

2. Learning Outcomes Analyze voltamogram and polarogram results.
By the end of the tutorial, student should be able to:
Analyze voltamogram and polarogram results.
Apply standard addition technique to voltametry and polarography.
Define cyclic voltammetry principles.
Analyze cyclic voltamogram
Apply CV principles on drug analysis.

3. Example I-
In 1 M NH3/1 M NH4Cl solution, Cu2+ is reduced to Cu+ near –0.3 V (versus SCE), and Cu+ is reduced to Cu (on Hg) near –0.6 V.
a) Sketch a qualitative sampled current polarogram for a solution of Cu+.
b) Sketch a polarogram for a solution of Cu2+.
c) Suppose that Pt, instead of Hg, were used as the working electrode. Which, if any, reduction potential would you expect to change?
Solution
(c) The potential for the reaction Cu(I) Cu(Hg) will change if Pt is used, since the product obviously cannot be copper amalgam and the potential at which reduction of Cu+ starts definitely depends on the type of electrode upon which redox reaction occurs.

4. Solution: Number of moles of electrons flowing in 3.4 min=
Example II-
Suppose that the diffusion current in a polarogram for reduction of Cd2+ at a mercury electrode is 14 A. if the solution contains 25 mL of 0.50 M Cd2+, what percentage of Cd2+ is reduced in the 3.4 min required to scan from 0.6 to 1.2 V? comment!
Solution: Number of moles of electrons flowing in 3.4 min=
For the reaction: Cd2+ + 2e Cd(Hg)
moles of Cd2+ = 1/2 moles of e- = 1.48 x 10-8 mol (forming amalgam)
moles of Cd2+ in 25 mL of 0.5 M solution = 25 x 0.50 x = 1.25 x 10-2 mol (starting concentration)
percentage Cd2+ of reduced =
Comment: In polarography, it is clear that a very small amount of Cd+2 (1.2 x 10-4%)has been reduced during the full time of the experiment(3.4 min). This is in contrast to electrogravimetry and coulometry in which all the analyte should be quantitatively reduced on the electrode surface. This little amount is representative of the concentration in builidng up a calibration curve.

5. Example III-
The polarogram of 3.00 mL of solution containing the antibiotic tetracycline in 0.1 M acetate, pH 4, gives a maximum current of 152 nA at a potential of –1.05 V (versus S.C.E.). When mL containing 2.65 mg/mL of tetracycline was added, the current increased to 206 nA. Calculate the concentration of tetracycline in the original solution.
Solution
In this case, both std. solution and unknown solution are mutually dilute each other.

6. Example IV- (Self Study)
The drug Librium gives a polarographic wave with E1/2 = 0.265 V (versus SCE) in 0.05 M H2SO4. A 50.0 mL sample containing Librium gave a wave height of 0.37 A. When 2.00 mL of 3.00 mM Librium in 0.05 M H2SO4 were added to the sample, the wave height increased to 0.80 A. Find the molarity of Librium in the unknown.
Answer: mM

7. Cyclic Voltammetry (CV) Irreversible reaction
forward
reversed
1st cycle
2nd cycle
Initial potential
final potential
Reversible reaction
Criteria for reversible reactions:
The ratio of the cathodic peak current (Ipc) to the anodic peak current (Ipa) should be 1.
The half-wave potential (E1/2) is the same
for the cathodic and anodic waves.
The separation between the cathodic peak potential
(Epc) and the anodic peak potential (Epa) should be close to 59/n mV
Irreversible reaction
Ep = |Epa - Epc| = 59/n mV

8. Example VI-
Consider the cyclic voltammogram of the Co3+ compound Co(B9C2H11)2-. Suggest a chemical reaction to account for each wave. Are the reaction reversible? How many electrons are involved in each step? Sketch the sampled current polarogram expected for this compound.
First wave
Second wave

9. Solution
We see two consecutive reactions.
From the value of Epa – Epc, we find that one electron is involved in each reduction (Ep = |Epa - Epc| = 59/n mV )
A possible sequence of reaction is:
Co Co2+
Co Co+
The equality of the anodic and cathodic heights suggests that the reactions are reversible (This is confirmed by the same value of half wave potential E1/2 in the cathodic and anodic directions).
The expected sampled current polarogram:
+ e-
+ e-
current E
voltage

10. Try to solve problems 17-22, 17-26, and 17-27
Example VII- A bonus idea
The cyclic voltammogram of the antibiotic chloramphenicol (abbreviated RNO2) is shown in the figure below. The scan was started at 0 V, and the potential was swept in the negative voltage. The first cathodic wave, A, is from the reaction
RNO2 + 4e- + 4H RNHOH + H2O.
Explain what happens at peaks B and C using the reaction
RNO + 2e- +2H RNHOH
Why was peak C not seen in the initial scan?
Answer
Peak B is due to the anodic reaction:
RNHOH RNO + 2e- +2H+
Peak C is due to the cathodic reaction:
RNO + 2e- +2H RNHOH
There was no peak C in the first run because no RNO intermediate B present before the initial scan.
Try to solve problems 17-22, 17-26, and 17-27
(Harris text book, p )